Question

The temperature at which a molecule of nitrogen will have the same rms velocity as a molecule of oxygen at $127^{\circ }C$ is

• A. $457℃$
• B. $273℃$
• C. $350℃$
• D. $77℃$

Answer 1

Answer: (D)

Root mean square velocity
$v_{rms}=\sqrt{\frac{3RT}{M}}$
where $R$ is gas constant, $T$ the temperature and $M$ molecular weight.
Given, $M_{N_2}=28,M_{O_2}=32,T_{O_2}=127{}^{\circ }C=127+273=400K$
$\therefore \frac{v_{O_2}}{v_{N_2}}=\sqrt{\frac{T_{O_2}}{M_{O_2}}\times \frac{M_{N_2}}{T_{N_2}}}=\sqrt{\frac{400}{32}\times \frac{28}{T_{N_2}}}=1$
$\Rightarrow T_{N_2}=350K=77{}^{\circ }C$